Problem: The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered. 

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 103 ≤ x, y ≤ 103), and r (1 ≤ r ≤ 103) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line. 

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10
60 0 10
30 30 10


Output

30.00000 0.00000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i];
    if (i + 1 != vec.size()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << itr->first << ":" << itr->second;
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << itr->first << ":" << itr->second;
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
using ll = long long int;
using P = pair<int, int>;
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
struct fast_io {
  fast_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(5);
  }
} fast_io_;
int main(int argc, char const *argv[]) {
  vector<long double> x(3), y(3), r(3);
  for (int i = (0); i < (3); ++i) cin >> x[i] >> y[i] >> r[i];
  if (r[0] == r[1] && r[1] == r[2]) {
    long double a1 = x[0] - x[1], a2 = x[1] - x[2];
    long double b1 = y[0] - y[1], b2 = y[1] - y[2];
    long double c1 = -(x[0] + x[1]) / 2. * (x[0] - x[1]);
    c1 -= (y[0] + y[1]) / 2. * (y[0] - y[1]);
    long double c2 = -(x[1] + x[2]) / 2. * (x[1] - x[2]);
    c2 -= (y[1] + y[2]) / 2. * (y[1] - y[2]);
    long double rem = a1 * b2 - a2 * b1;
    if (abs(rem) < EPS) {
      return 0;
    }
    long double X = (-b2 * c1 + b1 * c2) / rem;
    long double Y = (a2 * c1 - a1 * c2) / rem;
    cout << X << " " << Y << endl;
  } else {
    if (r[0] == r[1] || r[1] == r[2]) {
      swap(r[0], r[1]);
      swap(x[0], x[1]);
      swap(y[0], y[1]);
    }
    if (r[0] == r[1] || r[1] == r[2]) {
      swap(r[1], r[2]);
      swap(x[1], x[2]);
      swap(y[1], y[2]);
    }
    long double a1 = (r[0] * r[0] - r[1] * r[1]);
    long double b1 = -2. * (r[0] * r[0] * x[1] - r[1] * r[1] * x[0]);
    long double c1 = -2. * (r[0] * r[0] * y[1] - r[1] * r[1] * y[0]);
    long double d1 = -r[0] * r[0] * x[1] * x[1] - r[0] * r[0] * y[1] * y[1];
    d1 += r[1] * r[1] * x[0] * x[0] + r[1] * r[1] * y[0] * y[0];
    long double a2 = (r[1] * r[1] - r[2] * r[2]);
    long double b2 = -2. * (r[1] * r[1] * x[2] - r[2] * r[2] * x[1]);
    long double c2 = -2. * (r[1] * r[1] * y[2] - r[2] * r[2] * y[1]);
    long double d2 = -r[1] * r[1] * x[2] * x[2] - r[1] * r[1] * y[2] * y[2];
    d2 += r[2] * r[2] * x[1] * x[1] + r[2] * r[2] * y[1] * y[1];
    d1 += (b1 / (2. * a1)) * (b1 / (2.)) + (c1 / (2. * a1)) * (c1 / (2.));
    d2 += (b2 / (2. * a2)) * (b2 / (2.)) + (c2 / (2. * a2)) * (c2 / (2.));
    if (d2 / a2 < 0. || d1 / a1 < 0.) return 0;
    long double x0 = -b1 / (2. * a1), y0 = -c1 / (2. * a1), r0 = sqrt(d1 / a1);
    long double x1 = -b2 / (2. * a2), y1 = -c2 / (2. * a2), r1 = sqrt(d2 / a2);
    x1 -= x0, y1 -= y0;
    long double A = (x1 * x1 + y1 * y1 + r0 * r0 - r1 * r1) / 2.;
    if (x1 * x1 * r0 * r0 + y1 * y1 * r0 * r0 - A * A < 0.) return 0;
    long double X1 =
        (A * x1 - y1 * sqrt(x1 * x1 * r0 * r0 + y1 * y1 * r0 * r0 - A * A)) /
        (x1 * x1 + y1 * y1);
    long double Y1 =
        (A * y1 + x1 * sqrt(x1 * x1 * r0 * r0 + y1 * y1 * r0 * r0 - A * A)) /
        (x1 * x1 + y1 * y1);
    long double X2 =
        (A * x1 + y1 * sqrt(x1 * x1 * r0 * r0 + y1 * y1 * r0 * r0 - A * A)) /
        (x1 * x1 + y1 * y1);
    long double Y2 =
        (A * y1 - x1 * sqrt(x1 * x1 * r0 * r0 + y1 * y1 * r0 * r0 - A * A)) /
        (x1 * x1 + y1 * y1);
    X1 += x0, X2 += x0, Y1 += y0, Y2 += y0;
    long double t1 =
        (r[0] * r[0]) / ((X1 - x[0]) * (X1 - x[0]) + (Y1 - y[0]) * (Y1 - y[0]));
    long double t2 =
        (r[0] * r[0]) / ((X2 - x[0]) * (X2 - x[0]) + (Y2 - y[0]) * (Y2 - y[0]));
    if (t1 > t2)
      cout << X1 << " " << Y1 << endl;
    else
      cout << X2 << " " << Y2 << endl;
  }
  return 0;
}